if a voltage is applied to a capacitor, current flows easily at first and then slows as the capacitor becomes charged. Inductors
behave just the opposite, in that they reluctantly pass current when a voltage is first applied, and then the current passes easily as time passes. If the input voltage is suddenly raised from zero to some constant value, sketch the current in the capacitor, iC, and the inductor, iL, as a function of time. What is the steady-state current in the capacitor and inductor?
1 answer:
Answer: photo!
Explanation:
For the capacitor, the current as a function of time will be
I(t)=V/R e^(-t/RC)
In steady-state, time tends to infinite, so current tends to zero.
For the inductor, the current as a function of time is
I(t)=V/R (1-e^(-tR/L))
In steady-state, time tends to infinite, so current tends to V/R
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Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ